[SOLVED] OTA (opamp)out put impedance simulating

[Junus2012]

Hello all

I need kindly your help to give me an accurate reliable method method for simulating the output impedance of operation transconductance amplifier OTA,
I am usually using a common method by connecting a resistor at the output of an open loop op-ampand I do the d.C simulation and vary the resistor till I get half of the vo with no load and I consider this R as Ro, the problem with my simulation that I am getting a small Ro while it supposed to be in terms of Miga Ohms.
so what I did I changed the simulation to A.C simulation and I varied the RL till I got drop of the Av gain with -6db that is comparable to half of the gain with no load, Ro in this method was 7 MOhms that is reasonable, any way I didnt find any one simulating the output impedance by this way rather most people use the former method

that last method that I dont know how to use is by connecting an ideal current source at the output

Please I would like to know about the three of these simulation methods, and which one is more accurate

I appreciate too much any answer

 

[FvM]

To measure a resistance respecticely a complex impedance in a simulator, I'll inject a current into the node and measure the voltage.

But all other methods would work in AC analysis as well. The only important point is to achieve correct DC bias without introducing AC feedback or circuit loading.

 

[Junus2012]

Thank you for your reply

but could you please explain me more this method because this is the only one I am unable to use,

do you mean that I connect an ideal current source at the output , so what is the value of the current?? and the other setup condition ?

To measure a resistance respecticely a complex impedance in a simulator, I'll inject a current into the node and measure the voltage.

But all other methods would work in AC analysis as well. The only important point is to achieve correct DC bias without introducing AC feedback or circuit loading.

 

[FvM]

Stimulation current magnitude only matters for scaling in AC analysis. You can select any value.

The device under test must be biased correctly. In the example it's done by a huge time constant virtual LC filter.

 

[Junus2012]

Dear FvM

I did this simulation and tell me your opinion

I connected an a.c current source at the output of the amplifier with a magnitude of 1, in the same time I connected the normal signal voltage source that we use already when we simulate the frequency response of the amplifier. Then I run the A.C simulation, when I read the output voltage it will be automatically decided by the current source and give me the output impedance,
the simulation doesnt take any considerable time and it gave me the expected value of 7 Mohm as I have a cascoded output stage

What do you think about this method ( that I already read from an old post in this forum ), is right way??

 

[circuitrookie]

as long as you biased your circuit with dc voltage source and the only ac source is at the output then your method is fine

 

[Junus2012]

Thank you circuitrookie

I have two a.c sources, one is the voltage to generate Vo and the other one is the a.c current at the output node to have the division of Vo/Io. I dont know what you mean by only one a.c source ????how then you get V0. the other thing, of course I am biasing the circuit with the normal d.c biasing that I am using with all other kind of simulation

looking forward to your reply

as long as you biased your circuit with dc voltage source and the only ac source is at the output then your method is fine

 

[FvM]

in the same time I connected the normal signal voltage source that we use already when we simulate the frequency response of the amplifier.

If this means two AC sources active at the same time, it's clearly wrong.

 

[Junus2012]

Thank you again FvM

but tell me how then I can get Vo to be divided by the Io to get the output impedance ??-
and if we have to connect a.c current source then what will be the situation of the two input terminals ??

If this means two AC sources active at the same time, it's clearly wrong.

- - - Updated - - -

now I also believe that it is wrong to put two sources because the simulator will calculate it as Vo/Vi/I

 

[FvM]

because the simulator will calculate it as Vo/Vi/I

What should be the meaning of this expression? Vi hasn't to do with output impedance, Zo = Vo/Io
Io is injected in the test, Vo is the measured voltage.

 

[Junus2012]

Yes I am now totally agree with you but my question was what should be the status of the two input terminals of the op-amp, ok one is at VCM and what about the other ?? I dont have an inductor to use in my simulator as you show me in the image you have applied before.

looking forward to your reply

What should be the meaning of this expression? Vi hasn't to do with output impedance, Zo = Vo/Io
Io is injected in the test, Vo is the measured voltage.

 

[circuitrookie]

adding to FvM's response. if you check textbook of how output impedance is calculated in small signal model, you basically short out the input voltage source, so you don't need any other ac source to get Rout. i recommend you read about how to calculate output impedance from ss model then you would understand what you are doing in the simulation exactly.

 

[Junus2012]

Thank you circuitrookie for your response

As I already told FvM also, now I agree totally that I need one source in the circuit but what should be the case of the two input terminals of the op-amp, I am really sorry to ask this question again , and by the way when we are connecting the a.c current source at the output how then the frequency will affect the core amplifier ?

thank you again

adding to FvM's response. if you check textbook of how output impedance is calculated in small signal model, you basically short out the input voltage source, so you don't need any other ac source to get Rout. i recommend you read about how to calculate output impedance from ss model then you would understand what you are doing in the simulation exactly.

 

[FvM]

Yes I am now totally agree with you but my question was what should be the status of the two input terminals of the op-amp, ok one is at VCM and what about the other ??I dont have an inductor to use in my simulator as you show me in the image you have applied before.

The required connection can be derived from the definition of output impedance, or reviewd in an analog design text book, I presume.
- connect a DC input voltage that places the amplifier in linear operation range, e.g. output voltage at about mid supply
- no AC input

This is usually ahieved by DC feedback as in my example. I presume that a usable circuit simulator has some means to setup a DC bias circuit.

 

[Junus2012]

Ok, I will then connect the both terminals together to VCM where I can get Vo in the midway, and I think with this connection I dont need to use the feedback with my circuit, but the remaining thing how then I can provide the circuit with frequency change....????

I am really sorry to get long discussion with this topic
and I would thank you again

The required connection can be derived from the definition of output impedance, or reviewd in an analog design text book, I presume.
- connect a DC input voltage that places the amplifier in linear operation range, e.g. output voltage at about mid supply
- no AC input

This is usually ahieved by DC feedback as in my example. I presume that a usable circuit simulator has some means to setup a DC bias circuit.

 

[circuitrookie]

ac sim with your ac source will automatically find the output impedance at difference frequency. since it is a linear system. input frequency = output frequency

 

[Junus2012]

thank you all guys in advance for your helpful participation, the problem now are solved

see in the next problem

Have a nice day